Method for transmission of data packets through a network

ABSTRACT

The data packets with a maximum allowable delay and an average packet rate are aggregated in a node of the network of a burst and the burst is transmitted with a number of packets. The number of packets per burst is determined by the result of plus/minus twenty percent of the sum of the value one plus a product of the maximum allowable delay, the average packet rate and the value two.

CROSS REFERENCE TO RELATED APPLICATIONS

This application is the U.S. National Stage of International ApplicationNo. PCT/EP2004/051824, filed Aug. 18, 2004 and claims the benefitthereof. The International Application claims the benefits of Europeanapplication No. 03019475.7 EP filed Aug. 28, 2003, both of theapplications are incorporated by reference herein in their entirety.

FIELD OF INVENTION

The present invention relates to a method for transmission of datapackets through a network.

BACKGROUND OF INVENTION

In networks, like Optical Burst-Switched (OBS) networks or opticalnetworks, packets, e.g. Internet Protocol (IP) packets, AsynchronyTransfer Mode (ATM) cells or protocol data units (PDUs), are aggregatedto bursts, like electrical or optical bursts, in order to be transferredthrough the network. The conversion of packets into bursts takes placein a node, like an edge node, of the network according to a certainaggregation strategy. The solutions so far provide two main aggregationstrategies: the aggregation strategy with timeouts and the aggregationstrategy with buffer limit.

First we will discuss the aggregation strategy with timeouts. Aschematic example 100 is shown in FIG. 1. In this scheme, packets 102are added or padded to a burst 104 which is being generated in a buffer106 until a certain timer T expires. Then the burst 108 is sent.

The second aggregation strategy 200 with buffer limit will be discussedwith reference to FIG. 2. In this scheme, packets 202 are added orpadded to a burst 204 which is being generated in a buffer 206 until thebuffer is full. Then the burst 208 is sent.

Once the packets are transformed into bursts and sent into the network,they travel in the network through a series of switches, like electricalor optical switches, depending on the network, to a certain destination.At best, these switches have limited storage capabilities, e.g. in caseof optical switches fiber delay lines, and at worst, no storagecapabilities at all in the normal case. Therefore, collisions amongbursts occur. In a switch are two possible causes for the blocking of aburst.

First there is blocking due to switching time. A switch needs a certaintime t_(s) in order to process the header of an incoming burst and toprepare the switching elements so that the burst will be sent to theadequate output and channel, in case of optical switches to thecorresponding fiber and wavelength. This switching time t_(s) istechnology dependent. The blocking due to switching time occurs when aburst comes before the switching time t_(s) is elapsed.

Second there is blocking due to burst overlapping. In general, a switchhas a limited number of m channels per connection, in case of opticalswitches m wavelengths per optical fiber, going to a certaindestination. A burst is blocked when on a certain outgoing connectionall the m channels are busy, e.g. on a fiber all the m wavelengths arebusy.

On the other hand the burst generation time at the nodes is the maincause for the delay that a packet experiences. The burst generation timeis the time that a packet has to wait in the buffer at the node untilthe burst to which it belongs is completely generated.

SUMMARY OF INVENTION

Major performance parameters of a network are thus the burst blockingprobability, the throughput and the delay.

The two main aggregation strategies timeout and buffer limit have thedisadvantage of a certain blocking probability and an achievablethroughput.

It is an object of the invention to determine the number of packets perburst for a given maximum allowable delay of the packets and an averagepacket rate.

This object is achieved by the features recited in the claims.

The advantage of the invention is, that an optimum number of packets perburst is determined. In a network this number of packets per burst leadsto a lower blocking probability in the optical switches and to a higherthroughput.

Further developments of the invention are identified in the dependentclaims.

In an embodiment of the invention the number of packets per burst isused to determine the average burst size and the average burst rate.This has the advantage, that the parameters for calculating the trafficin the network are determined.

In an other embodiment of the invention the number of packets per burstis used to determine the relevant parameters for different aggregationstrategies, like buffer limit and random selection. This has theadvantage, that for every aggregation strategy the relevant parametersare determined to generate the optimum number of packets per burst, inorder to achieve a minimum blocking in the switches, to maximize thethroughput in the network and to remain below the maximum delay for apacket.

In an other embodiment of the invention the determined number of packetsper burst and out of this the resulting average burst rate is verifiedby an expression, using the inverse Erlang B formula for a given numberof m channels, a given burst blocking probability, a switching time anda burst duration time. This has the advantage, that the determinedresult for the number of packets per burst respectively the averageburst rate is verified by an independent formula.

The inventor found out, that for a given maximum allowable delay for anincoming packet and an average packet rate of the incoming packets theoptimum number of packets per burst ppb is given by plus/minus twentypercent of the sum of two terms, where the first term is the value oneand the second term is a product of the maximum allowable delay mad fora packet, the average packet rate apr and the value two.ppb=1+2·mad·apr  Equation 1

Beyond, for a transmission of data packets through a network, where saidpackets with a maximum allowable delay mad, an average packet rate aprand an average packet size aps are aggregated in a node of the networkto a burst and the burst is transmitted with an average number ofpackets and a link speed ls into the network and passes at least aswitch of the network, with a number of channels noc, a switching timet_(s) and a given burst blocking probability bbp, the average number ofpackets per burst ppb_(u) should be between an upper and a lower limit.

The upper limit of the average number of packets per burst ppbu shouldbe determined by plus/minus twenty percent of the sum of the value oneplus a product of the maximum allowable delay mad, the average packetrate apr and the value two.ppb _(u)=1+2·mad·apr  Equation 2

The lower limit of the average number of packets per burst ppbl shouldbe determined by plus/minus twenty percent of a quotient, where thedividend is the switching time t_(s) and the divisor is a difference,where the minuend is a quotient of the result of the inverse Erlang Bformula for the burst blocking probability bbp and the number ofchannels noc divided by the average packet rate apr and the subtrahendis a quotient of the average packet size aps divided by the link speedls. $\begin{matrix}{{ppb}_{l} = \frac{t_{s}}{\frac{E_{noc}^{- 1}({bbp})}{apr} - \frac{aps}{ls}}} & {{Equation}\quad 3}\end{matrix}$

Further on, an average burst size abs is given by the product of anaverage packet size aps of the incoming packets and the number ofpackets per burst ppb as determined above. The average packet size apscan be ascertained by measuring the size of the incoming packets anddetermine the average packet size aps. In case the incoming packets areIP packets, the results of the so called tri modal distribution for IPpackets can be used. According to this distribution the average packetsize aps of an IP packet is 3735 bits per IP packet.abs=aps·ppb  Equation 4

The determined number of the optimum average burst size abs, asdetermined by the product of the average packet size aps of the incomingpackets and the optimum number of packets per burst ppb as mentionedabove, can be used to generate bursts with the optimum size. During theaggregation of incoming packets to a burst an effective burst size, thismeans the size of a burst during aggregation at a certain time, iscompared to a value and if the effective burst size is equal or exceedsthis value the aggregated burst is sent and the incoming packets areaggregated to a new burst. This said value is in the range betweenplus/minus 20 percent of the optimum average burst size abs. The optimumaverage burst size should be used as preferred value.

Corresponding to the aggregation strategy with buffer limit, theincoming packets can be aggregated in a buffer to a bursts. According tothe above results, the minimum size of the buffer BSmin should be in therange determined by the product of the average packet size aps of theincoming packets and the difference of the number of packets per burstppb and a first constant d, which is greater than 0 and in maximum 5, apreferred value of 1 should be used.

The maximum size of the buffer BSmax should be determined by the productof two terms, where the first term is the average packet size aps andthe second term is a sum of the number of packets per burst ppb and asecond constant e, which is greater than 0 and in maximum 5, a preferredvalue of 1 should be used.BS min=aps·(ppb−d)  Equation 5BS max=aps·(ppb−e)  Equation 6

Commensurate to the aggregation strategy with random selection, asproposed in a parallel application of the patent applicant, every timean incoming packet is stored a random binary digit, with a probabilityfor the first and the second value of the binary digit, is generated andcompared with the first value of the binary digit. If both values areequal, the stored packets are sent as a burst. In order to generate withthis aggregation strategy the optimum number of packets per burst, theinventor proposes to determine the probability for the first value P(1)by the reciprocal value of the number of packets per burst ppb asdetermined above. The corresponding formula would be: $\begin{matrix}{{P(1)} = \frac{1}{ppb}} & {{Equation}\quad 7}\end{matrix}$

Anymore the average burst rate abr is determined by the quotient of theaverage packet rate apr and the number of packets per burst ppb.$\begin{matrix}{{abr} = \frac{apr}{ppb}} & {{Equation}\quad 8}\end{matrix}$

Alternatively the optimum average burst rate abr can be determined bythe reciprocal value of the sum of the inverse average packet rate aprand two times the maximum allowable delay mad for a packet.$\begin{matrix}{{abr} = \frac{1}{\frac{1}{apr} + {2 \cdot {mad}}}} & {{Equation}\quad 9}\end{matrix}$

A burst duration time t_(b) is the duration of a burst. Considering alink speed ls and the average burst size abs, the burst duration can bedetermined by the average burst size abs divided by the link speed ls.$\begin{matrix}{t_{b} = \frac{abs}{ls}} & {{Equation}\quad 10}\end{matrix}$

Considering a switching time t_(s), a burst duration time t_(b), a limitof the burst blocking probability bbp and a given number of channels nocin a switch of the network, the average burst rate abr should satisfythe condition that the quotient of the result of the inverse Erlang Bformula E_(noc) ⁻¹(bbp) for the limit of the burst blocking probabilitybbp and the number of channels noc divided by the sum of the switchingtime t_(s) and the burst duration time t_(b) is equal or greater thanthe average burst rate abr as determined before. $\begin{matrix}{{abr} \leq \frac{E_{noc}^{- 1}}{ls}} & {{Equation}\quad 11}\end{matrix}$

Where E_(noc) ⁻¹(bbp) respectively E_(m) ⁻¹(x) is the inverse Erlang Bformula for m channels, in case of OBS networks wavelengths, and for aload of x. In this case, the determined number of packets per burstrespectively the resulting average burst rate satisfies the condition,that the limit of the burst blocking probability for the givenparameters is not exceeded.

BRIEF DESCRIPTION OF THE DRAWINGS

An exemplary embodiment of the invention is described in greater detailbelow with reference to a drawing.

Shown in the drawing are:

FIG. 1 the initially cited prior art.

FIG. 2 the initially cited prior art.

FIG. 3 a diagram with the average burst rate as a function of theaverage burst size.

DETAILED DESCRIPTION OF INVENTION

FIG. 3 shows a diagram with three curves 10, 20, 30. On the x-axis theaverage burst size abs and on the y-axis the average burst rate abr isshown. Values 11, 12 for two average burst sizes are shown on the x-axisand a value 21 for an average burst rate is shown on the y-axis. Thefirst curve 10 is a throughput curve and displays a dependency betweenthe average burst size and the average burst rate for a given, constantthroughput TP, according the following formula:TP=abr·abs  Equation 12

When we move on the throughput curve to the left (smaller bursts andhigher rates, the burst blocking probability increases. The frequency ofbursts is higher, and therefore it is more probable that the switchingtime guard is not respected. The value 21 with the curve 20 imposes amaximum average burst rate for a given burst blocking probability. Thisleads two a maximum average burst rate and an average burst sizeaccording to the throughput curve, represented by a point A in FIG. 3.

The average number of packets per burst for point A is determined by aquotient, where the dividend is the product of the average packet rateapr and a sum of the switching time t_(s) and the burst duration timet_(b) and the divisor is the result of the inverse Erlang B formula forthe number of channels noc and the burst blocking probability bbp.$\begin{matrix}{{ppb}_{a} = \frac{{apr} \cdot ( {t_{s} + t_{b}} )}{E_{noc}^{- 1}({bbp})}} & {{Equation}\quad 13}\end{matrix}$

When we move on the throughput curve to the right (bigger bursts andlower rates, the delay increases. The bursts are bigger and therefore apacket has to wait in average longer until the burst to which it belongsis finished. The dependency of the average burst size and the averageburst rate for a given maximum allowable delay of a packet is shown bycurve 30 in FIG. 3. Concerning the throughput curve 10, for a giventhroughput and a given maximum allowable delay the maximum average burstsize and minimum average burst rate is represented by a point B in FIG.3.

In FIG. 3, the whole segment AB fulfills the condition of maximumallowable delay and given burst blocking probability. However, due tothe fact that B represents bigger burst sizes, this point B is betterthat any other point of the segment AB in the sense that it leads to ahigher multiplexing gain. For this reason, the desired optimum should bechosen near to the point B.

According to this, the above mentioned formulas determine an optimum ofthe average number of packets per burst.

For an edge node of an Optical Burst Switched (OBS) network, wherein onthe edge node incoming packets are aggregated to bursts, which are sentas optical bursts into the OBS network with a number of opticalswitches, the number of packets per burst will be determined by anexample.

An edge node receives traffic, e.g. IP packets from a STM-1 link with alink speed ls of 155,52 Mbits/s. The average load al of the link may be0,7. A tri modal distribution of the IP packets with an average IPpacket size aps of 3735 bits/packet is assumed. The average maximumallowable delay mad of an IP packet in the edge node should be below0,002 s.

We get the average packet rate:${apr} = {\frac{{ls} \cdot {al}}{aps} = {\frac{155,52\quad{Mbits}\text{/}{s \cdot 0},7}{3735\quad{bits}\text{/}{packet}} = {29146,988\quad{packets}\text{/}s}}}$

The optimal number of packets per burst ppb can be found with equation1, using c=1 and d=2:ppb=c+d·mad·apr=1+2·0.002 s·29146.988 packets/s=117.588 packets/burst

The optimal average burst size abs is then:abs=ppb·aps=117.588 packets/burst·3735 bits/packet=439191 bits/burst

The optimal average burst rate abr is:${abr} = {\frac{apr}{ppb} = {\frac{29146,988\quad{packets}\text{/}s}{117,588\quad{packets}\text{/}{burst}} = {247,\quad 873932\quad{bursts}\text{/}s}}}$

An OBS system with an optical switch receives traffic from the edgenode. The switching time t_(s) in the optical switch is 10 μs and thereare 8 wavelengths (channels) available per fiber. The burst blockingprobability bbp in the switch shall be kept below 10⁻⁶.

Using the inverse Erlang B formula for the number of channels noc and ablocking probability bbp of 10⁻⁶ we obtain:E₈ ⁻¹(10⁻⁶)=0.7337 erlang.

The average burst duration is:$t_{b} = {\frac{abs}{ls} = {\frac{439191\quad{bits}}{155,52\quad{bits}\text{/}s} = {2,\quad 8\quad{ms}}}}$

The condition:${{abr} \leq {\frac{E_{noc}^{- 1}({bbp})}{t_{s} + t_{b}}\text{=>}247,873932{{bursts}/s}} \leq \frac{0,7337{erlangs}}{{2,8{ms}} + {0,002s}}} = {259,715324{{burst}/s}}$is fulfilled, therefore the optimum calculated above satisfies the givenburst blocking probability.

It shall be appreciated that advantages of the invention are:

A higher throughput due to optimized burst sizes is achieved.

A given blocking probability is satisfied.

Due to given burst sizes, it is easier to calculate waiting times forbursts and/or headers of bursts.

The inventive method can be used to dynamically adapt the optimum numberof packets per burst for given or measured average packet rates and/oraverage packet sizes in an edge node of the network during operation. Sothe number of packets per burst can be adapted due to daily trafficfluctuations.

For example, the average rate of incoming packets is measured orotherwise ascertained. Corresponding to the maximum allowable delay andthe average packet rate the optimal number of packets is determined.According to this number and the used aggregation strategy, the optimalparameters for the concerning aggregation strategy are determined, e.g.for aggregation with buffer limit the requested buffer size, foraggregation with random selection the probability for the first value ofthe binary digit. With the inventive method these values can bedynamically readjusted, proportionately to the average incoming packetrate.

1.-15. (canceled)
 16. A method for transmitting data packets through acommunications network, the packets having a maximum allowable delay(mad), and average packet rate (apr) and an average packet size (aps),the method comprising: aggregating a plurality of data packets in a nodeof the network into a burst, the burst having an average number ofpackets; and transmitting the burst with a link speed (ls) into thenetwork via at least one switch having a number of channels (noc), aswitching time (ts) and a given burst blocking probability (bbp),wherein the average number of packets per burst (ppb) is between anupper and a lower limit, wherein the upper limit (ppb_(u)) is calculatedwith the formula:80%(1+2·mad·apr)≦ppb_(u)≦120%(1+2·mad·apr), and wherein the lower limit(ppb_(l)) is calculated with the formula:${80{\% \cdot \frac{t_{s}}{\frac{E_{noc}^{- 1}({bbp})}{apr} - \frac{aps}{ls}}}} \leq {ppb}_{l} \leq {120{\% \cdot \frac{t_{s}}{\frac{E_{noc}^{- 1}({bbp})}{apr} - \frac{aps}{ls}}}}$where E⁻¹ _(noc) is an inverse Erlang B formula for the noc.
 17. Themethod according to claim 16, wherein the number of packets per burst iswithin the lower limit.
 18. The method according to claim 16, whereinthe number of packets per burst is within the upper limit.
 19. Themethod according to claim 16, wherein the average burst size (abs) isabs=aps·ppb.
 20. The method according to claim 19, wherein during theaggregation of packets a burst size is compared to a value and if theburst size is equal to or exceed this value, the burst is transmitted,wherein the value is in the range between±20%·abs.
 21. The methodaccording to claim 19, wherein the packets are aggregated into a buffer,the minimum size of the buffer is determined by aps·(ppb−d), where d isa constant between 0 and
 5. 22. The method according to claim 21,wherein d is equal to
 1. 23. The method according to claim 19, whereinthe packets are aggregated into a buffer, the maximum size of the bufferis determined by aps·(ppb+e), where e is a constant between 0 and
 5. 24.The method according to claim 23, wherein e is equal to
 1. 25. Themethod according to claim 16, wherein when the packet is stored, arandom binary digit with a probability for a first and second value ofthe binary digit is generated and compared with the first value of thebinary digit, if equal the stored packets are sent as a burst, theprobability for the first value is determined by: $\frac{1}{ppb}.$ 26.The method according to claim 16, wherein an average burst rate isdetermined by: $\frac{apr}{ppb}.$
 27. The method according to claim 16,wherein an average burst rate is determined by:$\frac{1}{\frac{1}{apr} + {2 \cdot {mad}}}.$
 28. The method according toclaim 16, wherein the average burst rate(abr) is determined by:${abr} \leq \frac{E_{noc}^{- 1}({bbp})}{{ts} + {td}}$ where td is aburst duration time.
 29. The method according to claim 16, wherein thepackets are Internet Protocol (IP) packets.
 30. The method according toclaim 16, wherein the network is formed as an Optical Burst Switchednetwork
 31. The method according to claim 16, wherein the node is anedge node of the network
 32. The method according to claim 28, whereinthe burst is transformed and sent by the edge node as and optical burstinto the Optical Burst Switched network